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Polynomial of Degree 'n' =1,2,3,...

Which of the ...

Question

Which of the following represents a polynomial?

\frac{3}{x} + 2

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x^{2} + 3 x + \frac{4}{x}

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\sqrt{x} + 5

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3 x^{2} + \sqrt{5} x - \sqrt{6}

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Solution

The correct option is D $3 x^{2} + \sqrt{5} x - \sqrt{6}$
A polynomial of degree n is a function of the form-
$f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + . . . . . . a_{2} x^{2} + a_{1} x^{1} + x_{0}$
where the a’s are real numbers (sometimes called the coefficients of the polynomial).
$∴ 3 x^{2} + \sqrt{5} x - \sqrt{6}$ is a polynomial.

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Similar questions

Which of the following is a polynomial ?

(a) $x^{2} - 5 x + 4 \sqrt{x} + 3$ (b) $x^{\frac{3}{2}} - x + x^{\frac{1}{2}} + 1$

p (x)

g (x)

p (x) = x^{3} - 3 x^{2} + 5 x - 3, g (x) = x^{2} - 2

p (x) = x^{4} - 3 x^{2} + 4 x + 5, g (x) = x^{2} + 1 - x

p (x) = x^{4} - 5 x + 6, g (x) = 2 - x^{2}

x^{2} - 5 x + \sqrt{2}

4 x^{2} - 3 \sqrt{x} + 7

\frac{x^{2} + 2 x + 5}{x + 3} (x \neq - 3)

\frac{x^{3} - 5 \sqrt{x} - 1}{x^{2} + x + 4}

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

x^{2} + \frac{1}{x^{2}} = 5

x - \frac{3}{x} = x^{2}

2 x^{2} - \sqrt{3 x} + 9 = 0

x^{2} - 2 x - \sqrt{x} - 5 = 0

x + \frac{1}{x} = 1

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

x + \frac{1}{x} = x^{2}, x \neq 0

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